1
$\begingroup$

A Dedekind cut is a pair $(A,B)$ of nonempty subsets of $\mathbb Q$, satisfying:

  • $\mathbb Q = A \cup B$.

  • For all $a \in A$ and all $b\in B$, $a\le b$.

Define $\equiv$ by $(A_1,B_1) \equiv (A_2,B_2) \iff $ [for all $(a_1, b_2) \in A_1 \times B_2$, $a_1 \le b_2$ and for all $(a_2,b_1) \in A_2 \times B_1$, $a_2 \le b_1$]

Prove that $\equiv$ is an equivalence relation

Reflexivity and symmetry are obvious. For transitivity suppose that $(A_1,B_1) \equiv (A_2,B_2)$ and $(A_2,B_2) \equiv (A_3,B_3)$. We want to prove that $(A_1,B_1) \equiv (A_3,B_3)$.

Lemmas:

  • Let $(A,B)$ is a Dedekind cut. If $\max A$ exists then $A$ is $\{x \in \mathbb Q: x \le \max A\}$.

  • If $a,b \in \mathbb Q$ such that for all $0< \epsilon \in \mathbb Q$, $a \le b+\epsilon$, then $a \le b$.

Proof is easy.

Now take $a_1 \in A_1$ and $b_3 \in B_3$. If $b_3 \in B_2$, then we are done since $(A_1,B_1) \equiv (A_2,B_2)$. If not, then $b_3 \in A_2$, so since $a_2 \le b_3$ for all $a_2 \in A_2$, we get $A_2 = \{x \in \mathbb Q: x \le b_3\}$ and $B_2 = \{x \in \mathbb Q: x > b_3\}$ since $b_3 \notin B_2$. Then for every $0 < \epsilon \in \mathbb Q$, $b_3 + \epsilon \in B_2$ so $b_3 + \epsilon \ge a_1$ since $(A_1,B_1)\equiv (A_2,B_2)$, so we get $b_3 \ge a_1$. Similarly for the other one.

Is this OK? Is there another way to prove it? I want to know because in this paper that I am reading this is given as an exercise, the claim I made in the first point in the lemma uses in its proof something that is mentioned as the first part of the next theorem (on first page), so maybe I shouldn't have used it?

Link: http://spot.colorado.edu/~baggett/appendix.pdf

$\endgroup$
  • $\begingroup$ Your proof is fine, provided that you prove the parts of Theorem A.$1$. I don’t at the moment see a way to do it without using those two facts. $\endgroup$ – Brian M. Scott Sep 30 '16 at 22:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.